Ta có
$\sqrt{2018} - \sqrt{2017} = \dfrac{2018 - 2017}{\sqrt{2018} + \sqrt{2017}} = \dfrac{1}{\sqrt{2018} + \sqrt{2017}}$
$\sqrt{2016} - \sqrt{2015} = \dfrac{2016 - 2015}{\sqrt{2016} + \sqrt{2015}} = \dfrac{1}{\sqrt{2016} + \sqrt{2015}}$
Ta có
$\sqrt{2018} + \sqrt{2017} > \sqrt{2016} + \sqrt{2015}$
nên
$\dfrac{1}{\sqrt{2018} + \sqrt{2017}} < \dfrac{1}{\sqrt{2016} + \sqrt{2015}}$
Vậy
$\sqrt{2018} - \sqrt{2017} < \sqrt{2016} - \sqrt{2015}$.
